Theoretical DistributionsMCQPYQ Dec 21Question 3496 of 230
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The manufacturer of a certain electronic component is certain that 2%\displaystyle 2\% of his product is defective. He sells the components in boxes of 120 and guarantees that not more than 2%\displaystyle 2\% in any box will be defective. Find the probability that a box selected at random would fail to meet the guarantee? (e−2.4=0.0907\displaystyle e^{-2.4} = 0.0907)

Options

A0.49
B0.39
C0.37
D0.43
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Correct Answer

✅ Option d — 0.43

All Options:

  • A0.49
  • B0.39
  • C0.37
  • D0.43

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Detailed Solution & Explanation

Let n=120\displaystyle n = 120 be the number of components in a box, and let p=2%=0.02\displaystyle p = 2\% = 0.02 be the probability of any component being defective. Since n\displaystyle n is large and p\displaystyle p is small, we model the number of defective components in a box, X\displaystyle X, using a Poisson distribution with parameter: m=np=120×0.02=2.4m = np = 120 \times 0.02 = 2.4 The guarantee is that not more than 2%\displaystyle 2\% of the components in a box will be defective. 2% of 120=120×0.02=2.42\% \text{ of } 120 = 120 \times 0.02 = 2.4 Since the number of defectives must be an integer, the guarantee is met if the number of defectives is 0,1,\displaystyle 0, 1, or 2\displaystyle 2 (i.e., X≤2\displaystyle X \le 2). The box fails to meet the guarantee if X>2\displaystyle X > 2. We need to find P(X>2)\displaystyle P(X > 2): P(X>2)=1−P(X≤2)=1−[P(X=0)+P(X=1)+P(X=2)]P(X > 2) = 1 - P(X \le 2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)] The probability mass function is: P(X=k)=e−2.4(2.4)kk!P(X = k) = \frac{e^{-2.4} (2.4)^k}{k!} Using the given value e−2.4=0.0907\displaystyle e^{-2.4} = 0.0907: P(X=0)=e−2.4=0.0907P(X = 0) = e^{-2.4} = 0.0907 P(X=1)=2.4×e−2.4=2.4×0.0907=0.21768P(X = 1) = 2.4 \times e^{-2.4} = 2.4 \times 0.0907 = 0.21768 P(X=2)=(2.4)22!×e−2.4=2.88×0.0907=0.261216P(X = 2) = \frac{(2.4)^2}{2!} \times e^{-2.4} = 2.88 \times 0.0907 = 0.261216 Sum of probabilities for X≤2\displaystyle X \le 2: P(X≤2)=0.0907+0.21768+0.261216=0.569596P(X \le 2) = 0.0907 + 0.21768 + 0.261216 = 0.569596 Probability that the box fails the guarantee: P(X>2)=1−0.569596=0.430404≈0.43P(X > 2) = 1 - 0.569596 = 0.430404 \approx 0.43 Hence, **Option D** is the correct answer.

About This Chapter: Theoretical Distributions

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Binomial, Poisson, Normal Distribution

This chapter covers Binomial, Poisson, Normal Distribution and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

Key Concepts to Understand

Guarantee

A contract to perform the promise, or discharge the liability, of a third person in case of his default (Section 126, ICA 1872). It involves three parties: Principal Debtor, Creditor, and Surety. A guarantee is secondary liability.

Probability

A numerical measure of the likelihood of an event occurring, expressed as a number between 0 (impossible) and 1 (certain). Formula: P(A) = Number of favorable outcomes / Total number of equally likely outcomes.

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